Question

Customers enter the waiting line to pay for food as they leave a cafeteria on a first-come, first-served basis. The arrival rate follows a Poisson distribution, while service times follow an exponential distribution. If the average number of arrivals is four per minute and the average service rate of a single server is seven per minute, what proportion of the time is the serverbusy?A) 0.43B) 0.57C) 0.75D) 0.25E) 0.33

Answer 1

The answer is B) 0.57.

Explanation:

In this problem we have to apply queueing theory.

It is a single server queueing problem.

The arrival rate is
`\lambda=4` and the service rate is
`\mu=7`.

The proportion of time that the server is busy is now as the "server utilization"and can be calculated as:

`p=(\lambda)/(c\mu) =(4)/(1*7)=0.57`

where c is the number of server (in this case, one server).